IB Mathematics SL 1.9 The binomial theorem AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
General term in \( (x + h)^8 \): \( \binom{8}{r} \times x^{8-r} \times h^r \) (M1).
(i) For \( x^7 \): \( r = 1 \), term: \( \binom{8}{1} \times x^7 \times h^1 = 8 \times h \times x^7 \).
Thus, \( a = 8 \times h \) (A1 N1).
(ii) For \( x^6 \): \( r = 2 \), term: \( \binom{8}{2} \times x^6 \times h^2 = 28 \times h^2 \times x^6 \).
Thus, \( b = 28 \times h^2 \) (A1 N1).
(iii) For \( x^4 \): \( r = 4 \), term: \( \binom{8}{4} \times x^4 \times h^4 = 70 \times h^4 \times x^4 \).
Thus, \( d = 70 \times h^4 \) (A1 N1).
[3 marks]
Part (b)
Given: \( a, b, d \) form a geometric sequence, so \( \frac{b}{a} = \frac{d}{b} \).
Substitute: \( a = 8 \times h \), \( b = 28 \times h^2 \), \( d = 70 \times h^4 \) (M1).
Method 1: Direct Ratio Comparison
Set up: \( \frac{28 \times h^2}{8 \times h} = \frac{70 \times h^4}{28 \times h^2} \).
Simplify: \( \frac{28}{8} \times h = \frac{70}{28} \times h^2 \implies 3.5 \times h = 2.5 \times h^2 \) (A1).
Solve: \( 2.5 \times h^2 – 3.5 \times h = 0 \implies h \times (2.5 \times h – 3.5) = 0 \).
Since \( h > 0 \), \( 2.5 \times h = 3.5 \implies h = \frac{3.5}{2.5} = 1.4 \) (A1 A1 N2).
Method 2: Ratio Squared
Since \( \frac{b}{a} = \frac{d}{b} \), then \( \left(\frac{b}{a}\right)^2 = \frac{d}{a} \).
Calculate: \( \frac{b}{a} = \frac{28 \times h^2}{8 \times h} = 3.5 \times h \), so \( \left(3.5 \times h\right)^2 = \frac{70 \times h^4}{8 \times h} = 8.75 \times h^3 \) (M1).
Set up: \( (3.5 \times h)^2 = 8.75 \times h^3 \implies 12.25 \times h^2 = 8.75 \times h^3 \).
Solve: \( 8.75 \times h^3 – 12.25 \times h^2 = 0 \implies h^2 \times (8.75 \times h – 12.25) = 0 \).
Since \( h > 0 \), \( 8.75 \times h = 12.25 \implies h = \frac{12.25}{8.75} = 1.4 \) (A1 A1 N2).
Answer: \( h = 1.4 \) (A1).
[4 marks]
Total [7 marks]
▶️ Answer/Explanation
Part (a)
Expand \( (3 \times x + A) \times (2 \times x + B)^6 \). General term in \( (2 \times x + B)^6 \): \( \binom{6}{r} \times (2 \times x)^{6-r} \times B^r = \binom{6}{r} \times 2^{6-r} \times x^{6-r} \times B^r \).
Multiply by \( 3 \times x \) or \( A \) to get \( x^5 \)-terms (M1).
Case 1: \( 3 \times x \) term from first factor, \( x^4 \) from second
For \( x^4 \): \( 6-r = 4 \implies r = 2 \).
Term: \( \binom{6}{2} \times 2^{6-2} \times x^4 \times B^2 = 15 \times 2^4 \times x^4 \times B^2 \).
Multiply by \( 3 \times x \): \( 3 \times x \times 15 \times 2^4 \times x^4 \times B^2 = 3 \times 15 \times 16 \times B^2 \times x^5 = 720 \times B^2 \times x^5 \) (A1).
Case 2: \( A \) term from first factor, \( x^5 \) from second
For \( x^5 \): \( 6-r = 5 \implies r = 1 \).
Term: \( \binom{6}{1} \times 2^{6-1} \times x^5 \times B^1 = 6 \times 2^5 \times x^5 \times B \).
Multiply by \( A \): \( A \times 6 \times 2^5 \times x^5 \times B = 192 \times A \times B \times x^5 \) (A1).
Combine: \( (192 \times A \times B + 720 \times B^2) \times x^5 \).
Answer: \( \left( 192 \times A \times B + 720 \times B^2 \right) \times x^5 \) (A1 N2).
[4 marks]
Part (b)
Mina: \( P(\text{cookie}) = P(4, 5, \text{or } 6) = \frac{3}{6} = \frac{1}{2} \), \( P(\text{no cookie}) = \frac{1}{2} \).
Norbert: \( P(\text{cookie}) = P(5 \text{ or } 6) = \frac{2}{6} = \frac{1}{3} \), \( P(\text{no cookie}) = \frac{2}{3} \).
Exactly 5 cookies eaten: Mina eats 1 and Norbert eats 4, or Mina eats 0 and Norbert eats 5 (M1).
Method 1: Binomial Expansion
Relate to Part (a): Set \( A = P(\text{Mina cookie}) = \frac{1}{2} \), \( B = P(\text{Norbert cookie}) = \frac{1}{3} \), Norbert’s throws as \( (2 \times x + B)^6 \).
Coefficient of \( x^5 \): \( 192 \times \frac{1}{2} \times \frac{1}{3} + 720 \times \left(\frac{1}{3}\right)^2 = 192 \times \frac{1}{6} + 720 \times \frac{1}{9} = 32 + 80 = 112 \).
Probability: \( \frac{112}{6^6} = \frac{112}{46656} = \frac{4}{81} \approx 0.0494 \) (A1 A1 A1 N2).
Method 2: Direct Probability
Case 1: Mina eats 1, Norbert eats 4: \( P(\text{Mina 1}) = \frac{1}{2} \), \( P(\text{Norbert 4}) = \binom{6}{4} \times \left(\frac{1}{3}\right)^4 \times \left(\frac{2}{3}\right)^2 = 15 \times \frac{1}{81} \times \frac{4}{9} = \frac{60}{729} \).
Case 2: Mina eats 0, Norbert eats 5: \( P(\text{Mina 0}) = \frac{1}{2} \), \( P(\text{Norbert 5}) = \binom{6}{5} \times \left(\frac{1}{3}\right)^5 \times \frac{2}{3} = 6 \times \frac{1}{243} \times \frac{2}{3} = \frac{12}{729} \).
Total: \( \frac{1}{2} \times \frac{60}{729} + \frac{1}{2} \times \frac{12}{729} = \frac{30}{729} + \frac{6}{729} = \frac{36}{729} = \frac{4}{81} \approx 0.0494 \) (A1 A1 A1 N2).
Answer: \( \frac{4}{81} \) (A1).
[4 marks]
Total [8 marks]